Prove property of ranks

if A is an n x m matrix where n < m I would like to prove that there exists some [tex]\lambda[/tex] such that [tex]rank(A^T A + \lambda I) = m [/tex]

I know that if two of the columns of [tex]A^T A[/tex] are linearly dependent, they are scalar multiples of each other and by adding some [tex]\lambda[/tex] to two different positions, those colums will become independent but I can't prove it for more than two columns.

It seems to me that you're taking the pessimistic viewpoint. The rank will be full if the determinant is non-zero. The determinant of [tex]A^T A + \lambda I[/tex] is a polynomial in [tex]\lambda[/tex] , so will have non-zero determinant for all but finitely many [tex]\lambda[/tex].

I got the answer in simpler terms [tex]A^T A[/tex] is symmetric, so it is positive semi-definite and by taking any [tex]\lambda > 0[/tex] the matrix [tex]A^T A + \lambda I[/tex] is positive definite, hence non-singular and invertible.